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arjtryarjtry
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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 00:28
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In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)



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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 00:49
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should be D.

Not sure though
I am confident that B is sufficient on its own
2) 111111 is divisible by 7 and any other number will be x* 111111 ,where x is a digit other than 0.

1) The numbers i tried worked.
345345, 123123, 567567. will try to find whats the logic used here.
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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 00:50
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yes a+gamma ... the OA is D. good try... but will post the resoning later on. till then u can try
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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 01:05
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N = x x x x x x

F(k) = value of the kth digit.

From ( ii )

A B A B A B
will be divisible by 7.

From ( ii )

A A A A A A

will be divisible by 7


Thus answer should be "D"
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In a six-digit integer N F(k) is the value of the k-th Updated on: 29 Aug 2008, 08:41
Originally posted by x2suresh on 29 Aug 2008, 07:21.
Last edited by x2suresh on 29 Aug 2008, 08:41, edited 1 time in total.
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arjtryarjtry
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)
Show more

ABABAB
1) 100000*A +10000*B+ 1000*C +100*A + 10B +C
= 1001C +10010B+1001000A = 1001 (100A+10B+C)
= (100A+10B+C) * 1001
1001 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

100-2 --> 98 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.
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In a six-digit integer N F(k) is the value of the k-th Updated on: 29 Aug 2008, 08:35
Originally posted by x2suresh on 29 Aug 2008, 07:24.
Last edited by x2suresh on 29 Aug 2008, 08:35, edited 1 time in total.
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x2suresh
arjtryarjtry
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)

ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.
Show more

Hey guys,

I found another rule. (googled). this looks faster way for this problem

To know if a number is a multiple of seven or not, we can use also
3 coefficients (1 , 2 , 3). We multiply the first number starting
from the ones place by 1, then the second from the right by 3,
the third by 2, the fourth by -1, the fifth by -3, the sixth by -2,
and the seventh by 1, and so forth.

ABCABC --> 0 which is divisable by 7
AAAAAA -- > 0 which is divisable by 7
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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 08:30
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x2suresh
arjtryarjtry
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)

ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.
Show more

Am I reading this wrong? Is the number ABABAB or ABCABC? Obviously, your method would still be able to test:

ABCABC = 100000A + 10000B + 1000C + 100A + 10B + C = 100100A + 10010B + 1001C
= 1001*(100A + 10B + C)

1001/7 = 143 --> sufficient
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In a six-digit integer N F(k) is the value of the k-th 29 Aug 2008, 08:42
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zoinnk
x2suresh
arjtryarjtry
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)

ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.

Am I reading this wrong? Is the number ABABAB or ABCABC? Obviously, your method would still be able to test:

ABCABC = 100000A + 10000B + 1000C + 100A + 10B + C = 100100A + 10010B + 1001C
= 1001*(100A + 10B + C)

1001/7 = 143 --> sufficient
Show more

It should be ABCABC ... Sorry I did mistake.

I modified my origianl post.

Thanks



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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